Question

You are pushing a 20-kg box along a horizontal floor. Friction acts on the box. When you apply a horizontal force of magnitude 48 N, the box moves at a constant velocity. If you increase your horizontal pushing force to 100 N, what is the acceleration of the box?

Answer 1

Answer:

$a = 2.6\,\frac{m}{s^{2}}$

Explanation:

The first box has the following equation of equilibrium:

$\Sigma F = F - f = 0$

$f = F$

$f = 48\,N$

The coefficient of friction is:

$\mu_{k} = \frac{f}{m \cdot g}$

$\mu_{k} = \frac{48\,N}{(20\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}$

$\mu_{k} = 0.245$

There is a net acceleration, when horizontal pushing force is increased:

$\Sigma F = F - f = m\cdot a$

$a = \frac{F-f}{m}$

$a = \frac{100\,N-48\,N}{20\,kg}$

$a = 2.6\,\frac{m}{s^{2}}$

Answer 2

Explanation:

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